I know that there are several posts on the same idea, and I have read most of them, but still, my questions persist. I have listed the other posts on the topic at the end of this post. All the other posts say that the electric field inside an ideal wire must be zero because:

  1. As the potential drop across the wire is zero, the electric field must be zero.
  2. Since "J = σ E", and for an ideal wire the σ tends to infinity, so E must tend to zero.

I understand these two arguments, but still, I have the following questions.

Consider a simple circuit with a battery and a resistance. And, the two are connected with ideal wires.

The following are the questions related to it:

  1. If the electric field inside the ideal wires has to be zero, then how the current can flow in the circuit? Please explain to me with microscopic details, on how exactly the current flows in such a circuit and how exactly the electric field becomes zero inside the ideal wires.

  2. Is it the intrinsic property of a current-carrying ideal conductor itself that the electric field through it should be zero, or it manages to get electric field inside to be zero in all the scenarios? If an ideal conductor manages to get electric field inside it to be zero, in all the scenarios, then how exactly it does that?

I am expecting an explanation which can fit in the imagination as well. A more intuitive explanation.

Kindly help.

Following are some posts on similar ideas.

Is the electric field zero inside an ideal conductor carrying a current?

Is there a non-zero field within an ideal current carrying wire?


1 Answer 1


In a conductor the electric field applies a force to the conduction electrons so those electrons accelerate. The electrons then scatter off lattice vibrations (phonons) and decelerate. The current settles to an equilibrium state when the acceleration and deceleration have equal magnitudes, and when we do circuit analysis we assume that the circuit has settled to this equilibrium state. For example Ohm's law is only applicable at equilibrium.

The problem with an ideal conductor that has zero resistance is that there is no scattering of the electrons from phonons and hence no deceleration. That means if we were to apply a voltage across this conductor the electrons would just keep accelerating and the current would increase linearly with time without any upper limit. The circuit can never reach equilibrium and hence we cannot analyse it in the usual way.

The acceleration of the electrons happens inside the battery, or whatever power supply is being used. As each electron passes through the battery its kinetic energy is increased by $\Delta T = eV$, where $e$ is the electron charge and $V$ the battery voltage. Once the electrons leave the battery they are not subject to any field and just coast through the (ideal) conductor at constant velocity - a velocity that increases with each pass through the battery.

In a conductor with a resistance the potential difference appears because the electrons in the wire push on the electrons in front of them. The conductions electrons behave like a gas that it slightly more compressed at the entry to the wire than at the exit. If we use the hydraulic analogy we would say the pressure of the electron gas is higher at the start of the conductor than at the far end. The potential drop across the conductor is analogous to the pressure drop in the hydraulic analogy. However in a zero-resistance conductor there is no "pressure drop" because there is no resistance to the flow.

  • $\begingroup$ Thanks a lot for your response. But, my original questions still remain. 1. From your response I see two contradictory things coming together. First, the electrons accelerate inside an ideal wire, as there is no opposition and only acceleration due to electric field and in the second paragraph you also say that when electrons come into the idea wire, they are not subjected to any field. How do we reconcile these two statements? 2. My question is, whether this statement is true that in an ideal wire the electric field is zero? If yes then why? $\endgroup$ Jul 6, 2020 at 12:35
  • $\begingroup$ If there is electric field due to the battery inside the ideal wire then why do we say that it should be zero? What is making it zero inside the wire? $\endgroup$ Jul 6, 2020 at 12:36
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    $\begingroup$ @DevanshMittal Hmm, OK, yes, I've been a bit unclear. I need to go out now but I'll come back to this as soon as I'm back. $\endgroup$ Jul 6, 2020 at 12:36
  • $\begingroup$ Is it the case that there are two electric fields inside the wire - 1. Due to battery and 2. Due to the charge accumulation across the resistance. And, these two electric fields cancel inside the wire so net electric field inside the ideal wire becomes zero? Kindly confirm. $\endgroup$ Jul 6, 2020 at 12:37
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    $\begingroup$ @DevanshMittal I think it would be better to pursue this in the Physics chat room. I will be in that room for the next few hours. $\endgroup$ Jul 7, 2020 at 4:36

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